# Solve the following inequalities: 1/x + 2/(x-1)le 4

Solve the following inequalities:
$\frac{1}{x}+\frac{2}{x-1}\le 4$
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$\frac{1}{x}+\frac{2}{x-1}\le 4$
$\frac{1}{x}+\frac{2}{x-1}-4\le 0$
$\frac{1}{x}\ast \frac{\left(x-1\right)}{x-1}+\frac{2}{x-1}\ast \frac{x}{x}-4\ast \frac{x}{x}\ast \frac{x-1}{x-1}\le 0$
$\frac{x-1+2x-4{x}^{2}+4x}{x\left(x-1\right)}\le 0$
$\left(-1\right)\frac{-4{x}^{2}+7x-1}{x\left(x-1\right)}\le 0\left(-1\right)$
$\frac{4{x}^{2}-7x+1}{x\left(x-1\right)}\ge 0$
$4{x}^{2}-7x+1=0$
$x=\frac{-\left(-7\right)±\sqrt{\left(-7\right)\ast -4\left(4\right)\left(1\right)}}{2\left(4\right)}$
$=\frac{7±\sqrt{33}}{8}$
$x\left(x-1\right)=0$
$x=0;x=1$
$\sqrt{25}<\sqrt{33}<\sqrt{36}$
$5\sqrt{33}<6$
$\frac{7+\sqrt{33}}{8},\frac{7-\sqrt{33}}{8}$
$x=-1:\frac{4\left(-1{\right)}^{3}-7\left(-1\right)+1}{-1\left(-1-1\right)}=\frac{12}{2}\ge 0$
$x=0,001\approx 0:\frac{4\left(0.001{\right)}^{2}-7\left(0.001\right)+1}{0.001\left(0.001-1\right)}=\frac{1}{-\mathrm{#}}\text{⧸}>0$
$x=0.99:\frac{4\left(0.99{\right)}^{2}-7\left(0.99\right)+1}{0.99\left(0.99-1\right)}=\frac{-2}{-\mathrm{#}}\ge 0$
$x=1.001:\frac{4\left(1.001{\right)}^{2}-7\left(1.001\right)+1}{1.001\left(1.001-1\right)}=\frac{-2}{+\mathrm{#}}<0$
$x=10:\frac{4\left(10{\right)}^{2}-7\left(10\right)+1}{10\left(10-1\right)}=\frac{+}{+}\ge 0$
Solution: $\left(-\mathrm{\infty };0\right)\cup \left[\frac{7-\sqrt{33}}{8};1\right)\cup \left[\frac{7+\sqrt{33}}{8};\mathrm{\infty }\right]$
Jeffrey Jordon